Simulation process

The simple and fast way is that we simulate the whole process of policy and then find the best investment strategy with given upper bound on ruin probability. We assume we have known the estimated value of parameters in the model based on the real data. Then we do the four different experiments for Model 6 by using averaging process. For both experiments. We set one hundred different values of θ from 0 to 1. For every θ we use the value of parameters to simulate the value of surplus M times and record the value of surplus at the end of time. Finally we calculate the average value of it for every θ.

For experiment 3: Initial value A0=200, the standard deviation of the risk-investment interest rate σ1 = 0.15, the standard deviation of the bond interest rate σ2 = 0.1, the mean of claim size is 1, the term of policy n=60 and the recycle times M is 5000. For experiment 4, we keep the same value for all the parameters except we consider new σ1 =0.45.

The Figures 3.1 and 3.2 show that the average value of A60 against θ, the value of A60 is increasing when θ is increasing in both plots. In the Figure 3.1, the points distributed as a straight line, we believe the reason is that the value of σ1 = 0.15 is quite close to

σ2 = 0.1. However in the Figure 3.2, because σ1 = 0.45 is much bigger than σ2 = 0.1, so the points distributed as a curve. The value of A60on the Figure 3.2 is much bigger than on the Figure 3.1. The simulation tell us for the same investment strategy, the investments with high variance will take also high expected return.

Figure 3.1:Plot of the average value of An=60against θ (σ1=0.15, σ2=0.1)

Figure 3.3:Plot of the ruin probability for Model 6 against θ (σ1=0.15, σ2=0.1)

Figure 3.4:Plot of the ruin probability for Model 6 against θ (σ1=0.45, σ2=0.1)

In experiments 5 and 6, we still set one hundred values of θ from 0 to 1. For every θ, we use R language to generate M=5000 times value of Anand record how many times the simulated An < 0 in the Model 6. We end up the process with dividing this time described above by M, which is the simulated ruin probability we are looking for. For experiment 5, we take σ1 =0.15, σ2=0.1, A0=200 and n=60.

For experiment 6, we take σ1 =0.45, σ2=0.1, A0=200 and n=60.

The Figures 3.3 and 3.4 show the relationship between the simulated ruin probability and θ. Here we use the same value of parameters as in the first simulation. The surpris-

ing result is that there is a threshold in every plot which was empirically discovered. In the first plot, the threshold is around θ =0.5 compare with θ = 0.4 in the second plot. After the probability of ruin raise significantly. Another noticeable point is the value of simulated ruin probability on the Figure 3.4 plot is much higher than on the Figure 3.3 because of high value of σ1. The horizontal lines on both plots show A60=0.05 against different value of θ. It shows that the investment policy take more percentage of the first investment with higher variance by considering same ruin probability.

Discussion

In conclusion, the simulations support intuitive results. When the value of θ is increas- ing which means that the percentage of high risk investments is increasing, as a result, the ruin probability is increasing. However the rising of θ will make higher profit at time n=60. Therefore if there is an upper bound, on the ruin probability, for example 0.05 on the Figure 3.3 and 3.4, then the optimal investment policy is that the policy makes the ruin probability close to this value. Figure 3.3 shows that the θ as required is 0.93, then A60 with 0.93 on the Figure 3.1 is 298.6. Figure 3.4 shows that the θ we are looking for is 0.54, then A60with this value on the Figure 3.2 is 6641. This optimal policy is the balance, we are looking for, between the high return and the low risk. Now we extend our simulation process. For experiment 7, we still use the same initial value A0 = 200 and the variance of bond interest rate, σ22, is 0.01. Because σ1 should be bigger than σ2 , then we set one hundred different values of σ1 from 0.15 to 1.14. For every σ1, we find the optimal policy as described by considering an upper bound of ruin probability (0.5 here) above with the simulation. Figure 3.5 shows that the plot of the optimal θ against σ1. The value of θ is decreasing until σ1 is about 0.45. For

σ1 >0.45, θ appears to vary about the value value 0.5.

For every σ1, we have determined the optimal strategy, hence the optimal investment (surplus at time n) will be calculated. The Figure 3.6 is the plot of log(A60) against

σ1, and 1e+03 means 103on the figure. As we can see, the surplus is increasing when

Figure 3.5:Plot of the value of θ that gives the optimal policy against the standard de- viation of interest rate optimal (with σ2=0.1, n=60, A60=0.05,) experiment

7.

Figure 3.6:Plot of the standard deviation of the interest rate against the optimal in- vestment An(with σ2=0.1, A0=200, n=60,) experiment 7

Figure 3.7:Plot of the value of θ that gives the optimal policy against the standard de- viation of interest rate optimal (with σ2=0.1, n=60, A60=0.05,) experiment

8

Figure 3.8:Plot of the standard deviation of the interest rate against the optimal in- vestment An(with σ2=0.1, A0=200, n=60,) experiment 8

In experiment 8, for more realistic assumption, we set one hundred different values of

σ1from 0.10 to 0.19. And we repeat same simulation approach. The Figure 3.7 and 3.8 show the results of simulation. It is notable that there are two points on the Figure 3.7 close to 0. Because at that point, the value of σ1and σ2are quite close.